Let $(X, \mathcal{F},p)$ be a probability space and, for a fixed $A \in \mathcal{F}$, let $\chi_{A}(x)$ denote the characteristic function of the set $A$, that is, $\chi_{A}(x) = 1$ if $x \in A$ and zero otherwise. Let $\{f_{n}\}_{n\in \mathbb{N}}$ be a sequence of $L^{2}(X)$ functions such that $f_{n} \to \chi_{A}$ pointwise almost everywhere as $n \to \infty$. For each $n \in \mathbb{N}$, let $A_{n} = \{x \in X: \frac{1}{2} < f_{n}(x) \le 2\}$ and let $\chi_{A_{n}}$ be their associated characteristic function. I want to prove that $\chi_{A_{n}} \to \chi_{A}$ pointwise almost everywhere as $n \to \infty$.
My attempt: Since we are talking about "almost everywhere", let $M \subset X$ be the set of measure $p(M) = 1$ such that the pointwise convergence $f_{n}(x) \to \chi_{A}(x)$ occurs to every $x \in M$ as $n \to \infty$. Suppose $x \in M$ is such that $x \not\in A$ ($\chi_{A}(x) = 0$). In this case, $x \not\in A_{n}$ for sufficiently large $n$. In fact, if this was not the case, then for each $n_{0} \in \mathbb{N}$ there would exist an $n \ge n_{0}$ such that $x \in A_{n}$. But then, by the pointwise convergence hypothesis, there exists $n_{0} \in \mathbb{N}$ such that $|f_{n}(x) - \chi_{A}(x)| = |f_{n}(x)| < \frac{1}{2}$ for every $n \ge n_{0}$, which is a contradiction. Thus, $x \not\in A_{n}$ for all sufficiently large $n$ and $|\chi_{A_{n}}(x) - \chi_{A}(x)| = 0$ for $n \to \infty$. On the other hand, if $x \in M$ is such that $x \in A$, pick $n_{0} \in \mathbb{N}$ such that $\chi_{A}(x)-\frac{1}{2} = \frac{1}{2} < f_{n}(x) < \chi_{A}(x)+\frac{1}{2} = \frac{3}{2} < 2$. Then, for every $n \ge n_{0}$ we have $x \in A_{n}$ and $|\chi_{A_{n}}(x) - \chi_{A}(x)| = 0$.
I am not sure if my proof is correct or sufficiently rigorous. Is it correct? Can it be improved somehow? Maybe an easier way to prove it?
Why the contradiction argument? I think the conclusion can be seen directly. Let $x \in \{f_n \to \mathbf{1}_A\}$. For all $n$ sufficiently large, either $f_n(x)\in (-1/2,1/2)$ (if $x \in A^c$) or $f_n(x)\in (1/2,3/2)$ (if $x \in A$), so that respectively $\mathbf{1}_{A_n}(x)=0=\mathbf{1}_A(x)$ or $\mathbf{1}_{A_n}(x)=1=\mathbf{1}_A(x)$. So $1=p(f_n \to \mathbf{1}_A)\leq p(\mathbf{1}_{A_n}\to \mathbf{1}_{A})$.